POJ 2105 IP Address(水题)

题目链接

Description

Suppose you are reading byte streams from any device, representing IP addresses. Your task is to convert a 32 characters long sequence of '1s' and '0s' (bits) to a dotted decimal format. A dotted decimal format for an IP address is form by grouping 8 bits at a time and converting the binary representation to decimal representation. Any 8 bits is a valid part of an IP address. To convert binary numbers to decimal numbers remember that both are positional numerical systems, where the first 8 positions of the binary systems are: 
27   26  25  24  23   22  21  20 

128 64  32  16  8   4   2   1 

Input

The input will have a number N (1<=N<=9) in its first line representing the number of streams to convert. N lines will follow.

Output

The output must have N lines with a doted decimal IP address. A dotted decimal IP address is formed by grouping 8 bit at the time and converting the binary representation to decimal representation.

Sample Input

4
00000000000000000000000000000000 
00000011100000001111111111111111 
11001011100001001110010110000000 
01010000000100000000000000000001 

Sample Output

0.0.0.0
3.128.255.255
203.132.229.128
80.16.0.1
题意 :
把给出的32位2进制数每8位已转换,转换成4个十进制数,以IP地址的形式输出。
解题思路:
每8位一遍历相加。
/*
                                    _ooOoo_
                                   o8888888o
                                   88" . "88
                                   (| -_- |)
                                    O\ = /O
                                ____/`---'\____
                              .   ' \\| |// `.
                               / \\||| : |||// \
                             / _||||| -:- |||||- \
                               | | \\\ - /// | |
                             | \_| ''\---/'' | |
                              \ .-\__ `-` ___/-. /
                           ___`. .' /--.--\ `. . __
                        ."" '< `.___\_<|>_/___.' >'"".
                       | | : `- \`.;`\ _ /`;.`/ - ` : | |
                         \ \ `-. \_ __\ /__ _/ .-` / /
                 ======`-.____`-.___\_____/___.-`____.-'======
                                    `=---='

                 .............................................
*/
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define pi 3.1415926
#define INF 123456789
using namespace std;
int flag[8] = {1, 2, 4, 8, 16, 32, 64, 128};
int main(){
    int num1, num2, num3, num4, i, t;
    char str[100];
    scanf("%d%*c", &t);
    while(t --){
        scanf("%s", str);
        num1 = num2 = num3 = num4 = 0;
        for(i = 31; i >= 24; i --)
            if(str[i] == '1')
                num1 += flag[8 - (i - 24) - 1]; // 注意不要写成flag[8 - i - 24 - 1], 按照计算的优先级先算8-i为负数,所以要用括号括起来
        for(i = 23; i >= 16; i --)
            if(str[i] == '1')
                num2 += flag[8 - (i - 16) - 1];
        for(i = 15; i >= 8; i --)
            if(str[i] == '1')
                num3 += flag[8 - (i - 8) - 1];
        for(i = 7; i >= 0; i --)
            if(str[i] == '1')
                num4 += flag[8 - i - 1];
        printf("%d.%d.%d.%d\n", num4, num3, num2, num1);
    }
    return 0;
}
/*
*/


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